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Question Number 164533 by alephzero last updated on 18/Jan/22

$$\mathrm{Prove},\mathrm{that}$$$$1)\underset{k=1}{\sum ^{\alpha }}\frac{1}{k(k+1)}=\frac{\alpha }{\alpha +1}$$$$2)\underset{x\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k(k+x)}=0$$

Answered by amin96 last updated on 18/Jan/22

$$1)\underset{\mathbf{k}=1}{\sum ^{\alpha }}\frac{1}{\mathbf{k}}-\frac{1}{\mathbf{k}+1}=1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{\mathit{\alpha }}-(\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{\mathit{\alpha }+1})=$$$$=1-\frac{1}{\mathit{\alpha }+1}=\frac{\mathit{\alpha }}{\mathit{\alpha }+1}$$$$2)\underset{x\to \mathrm{\infty }}{\lim }(\frac{1}{\mathbf{x}}\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathbf{k}}-\frac{1}{(\mathbf{k}+\mathbf{x})})=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\mathbf{x}}\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathbf{k}}-\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\mathbf{x}}\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\frac{\frac{1}{\mathbf{x}}}{\frac{\mathbf{k}}{\mathbf{x}}+1}=$$$$=0-0=0$$

Commented by alephzero last updated on 18/Jan/22

$$\mathrm{Thank}\mathrm{You},\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 18/Jan/22

$$\frac{1}{k(k+1)}=\frac{A}{k}+\frac{B}{k+1}$$$$A(k+1)+Bk=1$$$$\{\begin{array}{l}k=0:A=1\\ k=-1:B=-1\end{array}$$$$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$$$$\underset{k=1}{\sum ^{\alpha }}\frac{1}{k(k+1)}=\underset{k=1}{\sum ^{\alpha }}(\frac{1}{k}-\frac{1}{k+1})$$$$\underset{k=1}{\sum ^{\alpha }}\frac{1}{k}-\underset{k=1}{\sum ^{\alpha }}\frac{1}{k+1}$$$$=\frac{1}{1}-\cancel{\frac{1}{2}}$$$$+\cancel{\frac{1}{2}}-\cancel{\frac{1}{3}}$$$$+\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}$$$$....$$$$+\cancel{\frac{1}{\alpha -1}}+\cancel{\frac{1}{\alpha }}$$$$+\cancel{\frac{1}{\alpha }}-\frac{1}{\alpha +1}$$$$=\frac{1}{1}-\frac{1}{\alpha +1}=\frac{\alpha +1-1}{\alpha +1}=\frac{\alpha }{\alpha +1}$$

Commented by alephzero last updated on 18/Jan/22

$$\mathrm{Thank}\mathrm{You}\mathrm{sir}.$$

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